I'm going through the question below.
The sequence [0, 1, ..., N] has been jumbled, and the only clue you have for its order is an array representing whether each number is larger or smaller than the last. Given this information, reconstruct an array that is consistent with it.
For example, given [None, +, +, -, +], you could return [1, 2, 3, 0, 4].
I went through the solution on this post but still unable to understand it as to why this solution works. I don't think I would be able to come up with the solution if I had this in front of me during an interview. Can anyone explain the intuition behind it? Thanks in advance!
This answer tries to give a general strategy to find an algorithm to tackle this type of problems. It is not trying to prove why the given solution is correct, but lying out a route towards such a solution.
A tried and tested way to tackle this kind of problem (actually a wide range of problems), is to start with small examples and work your way up. This works for puzzles, but even so for problems encountered in reality.
First, note that the question is formulated deliberately to not point you in the right direction too easily. It makes you think there is some magic involved. How can you reconstruct a list of N numbers given only the list of plusses and minuses?
Well, you can't. For 10 numbers, there are 10! = 3628800 possible permutations. And there are only 2⁹ = 512 possible lists of signs. It's a very huge difference. Most original lists will be completely different after reconstruction.
Here's an overview of how to approach the problem:
Start with very simple examples
Try to work your way up, adding a bit of complexity
If you see something that seems a dead end, try increasing complexity in another way; don't spend too much time with situations where you don't see progress
While exploring alternatives, revisit old dead ends, as you might have gained new insights
Try whether recursion could work:
given a solution for N, can we easily construct a solution for N+1?
or even better: given a solution for N, can we easily construct a solution for 2N?
Given a recursive solution, can it be converted to an iterative solution?
Does the algorithm do some repetitive work that can be postponed to the end?
....
So, let's start simple (writing 0 for the None at the start):
very short lists are easy to guess:
'0++' → 0 1 2 → clearly only one solution
'0--' → 2 1 0 → only one solution
'0-+' → 1 0 2 or 2 0 1 → hey, there is no unique outcome, though the question only asks for one of the possible outcomes
lists with only plusses:
'0++++++' → 0 1 2 3 4 5 6 → only possibility
lists with only minuses:
'0-------'→ 7 6 5 4 3 2 1 0 → only possibility
lists with one minus, the rest plusses:
'0-++++' → 1 0 2 3 4 5 or 5 0 1 2 3 4 or ...
'0+-+++' → 0 2 1 3 4 5 or 5 0 1 2 3 4 or ...
→ no very obvious pattern seem to emerge
maybe some recursion could help?
given a solution for N, appending one sign more?
appending a plus is easy: just repeat the solution and append the largest plus 1
appending a minus, after some thought: increase all the numbers by 1 and append a zero
→ hey, we have a working solution, but maybe not the most efficient one
the algorithm just appends to an existing list, no need to really write it recursively (although the idea is expressed recursively)
appending a plus can be improved, by storing the largest number in a variable so it doesn't need to be searched at every step; no further improvements seem necessary
appending a minus is more troublesome: the list needs to be traversed with each append
what if instead of appending a zero, we append -1, and do the adding at the end?
this clearly works when there is only one minus
when two minus signs are encountered, the first time append -1, the second time -2
→ hey, this works for any number of minuses encountered, just store its counter in a variable and sum with it at the end of the algorithm
This is in bird's eye view one possible route towards coming up with a solution. Many routes lead to Rome. Introducing negative numbers might seem tricky, but it is a logical conclusion after contemplating the recursive algorithm for a while.
It works because all changes are sequential, either adding one or subtracting one, starting both the increasing and the decreasing sequences from the same place. That guarantees we have a sequential list overall. For example, given the arbitrary
[None, +, -, +, +, -]
turned vertically for convenience, we can see
None 0
+ 1
- -1
+ 2
+ 3
- -2
Now just shift them up by two (to account for -2):
2 3 1 4 5 0
+ - + + -
Let's look at first to a solution which (I think) is easier to understand, formalize and demonstrate for correctness (but I will only explain it and not demonstrate in a formal way):
We name A[0..N] our input array (where A[k] is None if k = 0 and is + or - otherwise) and B[0..N] our output array (where B[k] is in the range [0, N] and all values are unique)
At first we see that our problem (find B such that B[k] > B[k-1] if A[k] == + and B[k] < B[k-1] if A[k] == -) is only a special case of another problem:
Find B such that B[k] == max(B[0..k]) if A[k] == + and B[k] == min(B[0..k]) if A[k] == -.
Which generalize from "A value must larger or smaller than the last" to "A value must be larger or smaller than everyone before it"
So a solution to this problem is a solution to the original one as well.
Now how do we approach this problem?
A greedy solution will be sufficient, indeed is easy to demonstrate that the value associated with the last + will be the biggest number in absolute (which is N), the one associated with the second last + will be the second biggest number in absolute (which is N-1) ecc...
And in the same time the value associated with the last - will be the smallest number in absolute (which is 0), the one associated with the second last - will be the second smallest (which is 1) ecc...
So we can start filling B from right to left remembering how many + we have seen (let's call this value X), how many - we have seen (let's call this value Y) and looking at what is the current symbol, if it is a + in B we put N-X and we increase X by 1 and if it is a - in B we put 0+Y and we increase Y by 1.
In the end we'll need to fill B[0] with the only remaining value which is equal to Y+1 and to N-X-1.
An interesting property of this solution is that if we look to only the values associated with a - they will be all the values from 0 to Y (where in this case Y is the total number of -) sorted in reverse order; if we look to only the values associated with a + they will be all the values from N-X to N (where in this case X is the total number of +) sorted and if we look at B[0] it will always be Y+1 and N-X-1 (which are equal).
So the - will have all the values strictly smaller than B[0] and reverse sorted and the + will have all the values strictly bigger than B[0] and sorted.
This property is the key to understand why the solution proposed here works:
It consider B[0] equals to 0 and than it fills B following the property, this isn't a solution because the values are not in the range [0, N], but it is possible with a simple translation to move the range and arriving to [0, N]
The idea is to produce a permutation of [0,1...N] which will follow the pattern of [+,-...]. There are many permutations which will be applicable, it isn't a single one. For instance, look the the example provided:
[None, +, +, -, +], you could return [1, 2, 3, 0, 4].
But you also could have returned other solutions, just as valid: [2,3,4,0,1], [0,3,4,1,2] are also solutions. The only concern is that you need to have the first number having at least two numbers above it for positions [1],[2], and leave one number in the end which is lower then the one before and after it.
So the question isn't finding the one and only pattern which is scrambled, but to produce any permutation which will work with these rules.
This algorithm answers two questions for the next member of the list: get a number who’s both higher/lower from previous - and get a number who hasn’t been used yet. It takes a starting point number and essentially create two lists: an ascending list for the ‘+’ and a descending list for the ‘-‘. This way we guarantee that the next member is higher/lower than the previous one (because it’s in fact higher/lower than all previous members, a stricter condition than the one required) and for the same reason we know this number wasn’t used before.
So the intuition of the referenced algorithm is to start with a referenced number and work your way through. Let's assume we start from 0. The first place we put 0+1, which is 1. we keep 0 as our lowest, 1 as the highest.
l[0] h[1] list[1]
the next symbol is '+' so we take the highest number and raise it by one to 2, and update both the list with a new member and the highest number.
l[0] h[2] list [1,2]
The next symbol is '+' again, and so:
l[0] h[3] list [1,2,3]
The next symbol is '-' and so we have to put in our 0. Note that if the next symbol will be - we will have to stop, since we have no lower to produce.
l[0] h[3] list [1,2,3,0]
Luckily for us, we've chosen well and the last symbol is '+', so we can put our 4 and call is a day.
l[0] h[4] list [1,2,3,0,4]
This is not necessarily the smartest solution, as it can never know if the original number will solve the sequence, and always progresses by 1. That means that for some patterns [+,-...] it will not be able to find a solution. But for the pattern provided it works well with 0 as the initial starting point. If we chose the number 1 is would also work and produce [2,3,4,0,1], but for 2 and above it will fail. It will never produce the solution [0,3,4,1,2].
I hope this helps understanding the approach.
This is not an explanation for the question put forward by OP.
Just want to share a possible approach.
Given: N = 7
Index: 0 1 2 3 4 5 6 7
Pattern: X + - + - + - + //X = None
Go from 0 to N
[1] fill all '-' starting from right going left.
Index: 0 1 2 3 4 5 6 7
Pattern: X + - + - + - + //X = None
Answer: 2 1 0
[2] fill all the vacant places i.e [X & +] starting from left going right.
Index: 0 1 2 3 4 5 6 7
Pattern: X + - + - + - + //X = None
Answer: 3 4 5 6 7
Final:
Pattern: X + - + - + - + //X = None
Answer: 3 4 2 5 1 6 0 7
My answer definitely is too late for your problem but if you need a simple proof, you probably would like to read it:
+min_last or min_so_far is a decreasing value starting from 0.
+max_last or max_so_far is an increasing value starting from 0.
In the input, each value is either "+" or "-" and for each increase the value of max_so_far or decrease the value of min_so_far by one respectively, excluding the first one which is None. So, abs(min_so_far, max_so_far) is exactly equal to N, right? But because you need the range [0, n] but max_so_far and min_so_far now are equal to the number of "+"s and "-"s with the intersection part with the range [0, n] being [0, max_so_far], what you need to do is to pad it the value equal to min_so_far for the final solution (because min_so_far <= 0 so you need to take each value of the current answer to subtract by min_so_far or add by abs(min_so_far)).
I was trying to recursively pass a function to itself a given number of times. So given the input in form of Function | RepeatNumber(Count) | Argument. So as an example if given the input: f 3 2 it would return f(f(f 2) This should then also work if the Function value where "square" it would Square the argument.
In terms of the logic I approached it as follows:
def repeatnew func count arg
if count == 1 then (func arg)
else (func (repeatnew func (count -1) arg))
I've been trying to research a solution to this for a while and I came across using Iterate and some other functions. Finally I came across this: https://wiki.haskell.org/Higher_order_function However I was unable to implement a working solution.
Edit: The solutions I did try to implement I could get compiling correctly, I am still very inexperienced with haskell and would appreciate an explanation on how to create a higher order function using my parameters.
Spinning off from #Antisthenes' comment, One other way to do this is using the foldl1 with no base case.
pipeNTimes :: (a -> a) -> Int -> (a -> a)
pipeNTimes f n = foldl1 (.) $ replicate n f
pipeNTimes (*2) 3 2 -- returns 16
I am admittedly only a beginner at Haskell, so this may be a naive implementation, but I think this does what you are looking for:
applyRecursively f x y
| x == 1 = f y
| otherwise = f (applyRecursively f (x-1) y)
The applyRecursively function takes a function and two numbers as arguments.
If the middle argument (the count) is 1 then we apply the parameter function to the argument y. This is the edge case that stops the recursion.
The otherwise guard clause then calls recursively with the x parameter decremented, and applies function f to the result.
I have tested it using a lambda like this:
applyRecursively (\x -> x + 1) 3 3
It should add 1 to the values of 3 three times - it returns a value of 6 so it looks like it works.
As I say, I'm a beginner but I think this does what you're looking for.
This question already has answers here:
Subscript indices must either be real positive integers or logicals, generic solution
(3 answers)
Closed 6 years ago.
Not sure if I should have posted this here or on the Maths stackexchange, so sorry if it's the wrong place. I'm very new to MATLAB and programming in general, and am having some troubles trying to solve an ODE problem using finite difference methods for an assignment.
My finite difference equation is:
z(t+dt) = (dt^2*(γ^2*h*sin(γ*t)-β*z(t)) - z(t-dt)*(1-dt*α)+2*z(t))/(1 + dt*α)
Where t is a 51x1 array for the time increments. Basically I want to calculate z(t) for t values from 0 to 1 in increments of 0.02. I have the initial conditions z(0) = 0 and z(Δt) = 0.
My current code (not everything, but the bit that's giving me trouble:
dt = 0.02
t = [0:dt:T]';
z(0) = 0
z(dt)= 0
for i = t
z(i+dt) = (dt^2*(gamma^2.*h.*sin(gamma*t)-beta*z(i)) - z(i-dt)*(1-dt*alpha)+2*z(i))/(1 + dt*alpha)
end
Alpha, beta and gamma are all constants in this case, they're defined earlier in the code.
I keep getting the error "Subscript indices must either be real positive integers or logicals." I understand that MATLAB arrays begin with element 1 and not 0, so trying to access element 0 will give an error.
I'm not sure if the error is with how I've entered my finite difference function, or the initial conditions. By setting i = t, am I running the for loop for those values of t, or for those elements in the matrix? E.g. when i = 0, is it trying to access the 0 element of the matrix, or is it setting the i variable in the equation to 0 like I want it to?
Any help would be greatly appreciated.
Thankyou!
problem is with dt and I think dt is not integer value and cannot be used for indexing array. index of arrays are always integer values in Matlab. Please try below code and check if solves the problem
t = [0:dt:T]';
z(1) = 0
z(2)= 0
for i = 2 : length(t)
z(i) = (dt^2*(gamma^2.*h.*sin(gamma*t(i))-beta*z(i)) - z(i-1)*(1-dt*alpha)+2*z(i))/(1 + dt*alpha)
end
Foremost, I saw this question before posting, mine is not a duplicate, our questions are different beside the titles.
I don't have a big knowledge in SQLSERVER/T-SQL, so when I was "translating" some scripts from ORACLE/PLSQL, I didn't found any function to generate a random number between 2 values.
In the PLSQL script I'm looking at, I can get it simply using DBMS_RANDOM.VALUE(5, 10).
How can I do that in T-SQL? Thanks!
You would use the rand function, which gives you a value 0 <= n < 1. Example for a number between 5 and 10 (inclusive):
floor(rand() * (10 - 5 + 1)) + 5
I am having trouble with a particular question as follows,
n is an exact power of 3
(n=3^k , where k is a positive integer)
while n > 1
begin
print "hello"
n := n/3
end
I need to find a function of n that will determine how many times 'hello' will be printed.
I am having trouble with it because of the k variable. Is this suppose to be implied within the equation. I realize some test data is n=3 -> 1 ; n=9 -> 2 ; n=27 -> 3 , but it just isn't clicking on how to incorporate k into the solution, if at all. Any push in the right direction would be greatly appreciated. Thanks
yes this is schoolwork, practice problems, but I am not looking for someone to just give me an answer I want to understand how to arrive at it.
Hint: if you have something to exponentiate, what is the reverse
operation?
Logarithms are your friends. =)